Let $f(x)=3x^4+4x^3-12x^2+5$. What is the absolute minimum value of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-27$ (Choice B) B $-54$ (Choice C) C $-100$ (Choice D) D $f$ has no minimum value
Let's first find the relative extremum points of $f$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $f$. The derivative of $f$ is $f'(x)=12x(x+2)(x-1)$. $f'(x)=0$ for $x=-2,0,1$. $f'$ is defined for all real numbers. Therefore, our critical points are $x=-2$, $x=0$, and $x=1$. Our critical points divide the function's domain (which is all real numbers) into four intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $(-\infty, \llap{-}2)$ $( \llap{-}2,0)$ $(0,1)$ $(1,\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,-2)$ $x=-3$ $f'(-3)=-144<0$ $f$ is decreasing $\searrow$ $(-2,0)$ $x=-1$ $f'(-1)=24>0$ $f$ is increasing $\nearrow$ $(0,1)$ $x=\dfrac{1}{2}$ $f'\left(\dfrac{1}{2}\right)=-7.5<0$ $f$ is decreasing $\searrow$ $(1,\infty)$ $x=2$ $f'(2)=96>0$ $f$ is increasing $\nearrow$ Now let's look at all the critical points: $x$ $f(x)$ Before After Verdict $-2$ $-27$ $\searrow$ $\nearrow$ Minimum $0$ $5$ $\nearrow$ $\searrow$ Maximum $1$ $0$ $\searrow$ $\nearrow$ Minimum Let's imagine ourselves walking on the graph of $f$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until $(-2,-27)$, then go up until $(0,5)$, then down again until $(1,0)$, and then forever go up. This means that $\lim_{x\to-\infty}f(x)=\lim_{x\to +\infty}f(x)=+\infty$, which means $f$ has no maximum value. However, $f$ does reach an absolute minimum point at $(-2,-27)$, which means its absolute minimum value is $-27$. In conclusion, the absolute minimum value of $f$ is $-27$.